# Problem of the Week #91 - February 24th, 2014

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#### Chris L T521

##### Well-known member
Staff member
Here's this week's problem!

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Problem
: Suppose that $A_1\supseteq A_2\supseteq A_3\supseteq\cdots\supseteq A_n\cdots$ is a sequence of measurable sets with $m\left(A_1\right)<\infty$. Show that $$m\left( \bigcap\limits_{i=1}^{\infty}A_i\right) =\lim\limits_{i\to\infty}m\left(A_i\right).$$

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#### Chris L T521

##### Well-known member
Staff member
This week's problem was correctly answered by Ackbach (quoting Royden as his source). You can find his solution below.

I freely admit that I am essentially stealing this solution lock, stock, and barrel from Royden's Real Analysis, page 63. I'm only changing the notation, since I despise Royden's set notation. His setminus is a $\sim$, and he notates the measure of $A$ by $mA$. I will use $-$ for setminus, and the measure of $A$ by $m(A)$.

Let
$$\displaystyle A=\bigcap_{i=1}^{\infty}A_{i},$$
and let $B_{i}=A_{i} - A_{i+1}$. Then
$$A_{1} - A = \bigcup_{i=1}^{\infty} B_{i},$$
and the sets $B_{i}$ are pairwise disjoint.
Hence
$$m(A_{1} - A)= \sum_{i=1}^{\infty}m(B_{i})= \sum_{i=1}^{\infty}m(A_{i} - A_{i+1}).$$

But $m(A_{1})=m(A)+m(A_{1} - A),$ and $m(A_{i})=m(A_{i+1})+m(A_{i} - A_{i+1}),$ since $A \subseteq A_{1}$ and $A_{i+1} \subseteq A_{i}$.
Since $m(A_{i}) \le m(A_{1}) < \infty,$ we have $m(A_1 - A)=m(A_1)-m(A)$ and
$m(A_1 - A_{i+1})=m(A_i)-m(A_{i+1}).$

Thus

\begin{align*}
m(A_1)-m(A)&= \sum_{i=1}^{\infty}[m(A_i)-m(A_{i+1})] \\
&= \lim_{n \to \infty} \sum_{i=1}^{n}[m(A_i)-m(A_{i+1})] \\
&= \lim_{n \to \infty}[m(A_1)-m(A_n)] \\
&=m(A_1)- \lim_{n \to \infty}m(A_n).
\end{align*}

Since $m(A_1)< \infty,$ we have $\displaystyle m(A)= \lim_{n \to \infty}m(A_n).$

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